Question

A laboratory furnace wall is constructed of 0.2 m thick fireclay brick having a thermal conductivity of 1.82 W/m-K. The wall is covered on the outer surface with insulation of thermal conductivity of 0.095 W/m-K. The furnace inner brick surface is at 950 K and the outer surface of the insulation material is at 300 K. The maximum allowable heat transfer rate through the wall of the furnace is 830 W/m^2. Determine how thick in cm the insulation material must be.

Answer 1

The appropriate solution will be "6.4 cm".

Step-by-step explanation:

The given values are:

Length,

l = 0.2 m

Thermal conductivity,

K₁ = 1.82 W/m-K

K₂ = 0.095 W/m-K

Temperature,

T = 950 K

T = 300 K

Heat transfer rate,

Q = 830 W/m²

Now,

⇒
`Q = (\Delta T)/((L_1)/(K_1 A) +(L_2)/(K_2 A) )=(A \Delta T)/((L_1)/(K_1 ) +(L_2)/(K_2 ) )`

⇒
`(Q)/(A) =(\Delta T)/((L_1)/(K_1) +(L_2)/(K_2) )`

On substituting the above given values in the equation, we get

⇒
`830=((980-300))/((0.2)/(1.82) +(x)/(0.095) )`

On applying cross-multiplication, we get

⇒
`(0.2)/(1.82) +(x)/(0.095) =(950-300)/(830)`

⇒
`(0.2)/(1.82) +(x)/(0.095) =(650)/(830)`

⇒
`x =0.639 \ m`

⇒
`x=6.345 \ i.e., 6.4 \ m`